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insens350 [35]
2 years ago
15

The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of which fun

ction(s)? tangent only cosine only sine and tangent cosine and sine
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
4 0

The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

<h3>What is vertical motion of a projecile?</h3>

The vertical motion of a projectile is affected by gravity and the velocity of vertical motion given by the following formula;

Vy = Vsinθ

<h3>What is horizontal motion of a projecile?</h3>

The horizontal velocity of a projectile is given by the following formula;

Vx = Vcosθ

<h3>Direction of the motion</h3>

The direction of the motion is calculated as follows;

tanθ  = Vy/Vx

Thus, the formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

Learn more about vertical motion here: brainly.com/question/24216590

#SPJ4

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3 years ago
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kobusy [5.1K]

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3 0
2 years ago
The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune
Veronika [31]
Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
\frac{r^3}{T^2}= \frac{GM}{4 \pi^2} (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
1 AU = 1.5 \cdot 10^{11} m
so the radius of the orbit is
r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
T=\sqrt{ \frac{4 \pi^2}{GM} r^3} =  \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years
3 0
3 years ago
Can anyone help me with questions a and c​
Ganezh [65]

Answer:

a)

n=sin i/sin r

n= -0.305/-0.428

n=0.713

b)

sin c=1/n

sin c=1/0.713

sin c= 1.403

c=sin⁻¹(1.403)

c= 40.842°

Explanation:

i hope it will be helpful

plzzz mark as brainliest

5 0
3 years ago
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